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Can Someone Answer My Final Answers :) I’d appreciate it so much :)

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User Wayne Wang
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Hello,
Please, see the attached files.
Thanks.
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User Jason Rahm
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1. Remember that the perimeter is the sum of the lengths of the sides of a figure.To solve this, we are going to use the distance formula:
d= \sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2}
where

(x_(1),y_(1)) are the coordinates of the first point

(x_(2),y_(2)) are the coordinates of the second point
Length of WZ:
We know form our graph that the coordinates of our first point, W, are (1,0) and the coordinates of the second point, Z, are (4,2). Using the distance formula:

d_(WZ)= √((4-1)^2+(2-0)^2)

d_(WZ)= √((3)^2+(2)^2)

d_(WZ)= √(9+4)

d_(WZ)= √(13)

We know that all the sides of a rhombus have the same length, so

d_(YZ)= √(13)

d_(XY)= √(13)

d_(XW)= √(13)

Now, we just need to add the four sides to get the perimeter of our rhombus:

perimeter= √(13) + √(13) + √(13) + √(13)

perimeter=4 √(13)
We can conclude that the perimeter of our rhombus is
4 √(13) square units.

2. To solve this, we are going to use the arc length formula:
s=r \alpha
where

s is the length of the arc.

r is the radius of the circle.

\alpha is the central angle in radians

We know form our problem that the length of arc PQ is
(8)/(3) \pi inches, so
s=(8)/(3) \pi, and we can infer from our picture that
r=15. Lest replace the values in our formula to find the central angle POQ:

s=r \alpha

(8)/(3) \pi=15 \alpha

\alpha = ((8)/(3) \pi)/(15)

\alpha = (8)/(45) \pi

Since
\alpha =POQ, We can conclude that the measure of the central angle POQ is
(8)/(45) \pi

3. A cross section is the shape you get when you make a cut thought a 3 dimensional figure. A rectangular cross section is a cross section in the shape of a rectangle. To get a rectangular cross section of a particular 3 dimensional figure, you need to cut in an specific way. For example, a rectangular pyramid cut by a plane parallel to its base, will always give us a rectangular cross section.
We can conclude that the draw of our cross section is:
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User Ironsand
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